NOI2002 荒岛野人
这题其实黑书上有,只是我脑残的没想起来……
其实就是拓展欧几里得算法
我参看的题解:http://www.cnblogs.com/Rinyo/archive/2012/11/25/2788373.html
还有一个讲解的很清楚的exgcd:http://www.cnblogs.com/Rinyo/archive/2012/11/25/2787419.html
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 var c,p,l:array[0..50] of longint; 2 ans,now,n,i,j,d,x,y,pp,cc:longint; 3 flag:boolean; 4 function min(x,y:longint):longint; 5 begin 6 if x<y then exit(x) else exit(y); 7 end; 8 function max(x,y:longint):longint; 9 begin 10 if x>y then exit(x) else exit(y); 11 end; 12 function exgcd(a,b:longint;var x,y:longint):longint; 13 var t,d:longint; 14 begin 15 if b=0 then 16 begin 17 x:=1;y:=0;exit(a); 18 end; 19 d:=exgcd(b,a mod b,x,y); 20 t:=x;x:=y;y:=t-(a div b)*y; 21 exit(d); 22 end; 23 procedure main; 24 begin 25 readln(n); 26 for i:=1 to n do 27 begin 28 readln(c[i],p[i],l[i]); 29 ans:=max(ans,c[i]); 30 end; 31 while true do 32 begin 33 flag:=true; 34 for i:=1 to n-1 do 35 begin 36 for j:=i+1 to n do 37 begin 38 pp:=p[i]-p[j];cc:=c[j]-c[i]; 39 d:=exgcd(pp,ans,x,y); 40 if cc mod d<>0 then continue; 41 now:=x*cc div d; 42 now:=now mod (ans div d); 43 if now<0 then inc(now,abs(ans div d)); 44 if now<=min(l[i],l[j]) then begin flag:=false;break;end; 45 end; 46 if not(flag) then break; 47 end; 48 if flag then break; 49 inc(ans); 50 end; 51 writeln(ans); 52 end; 53 begin 54 main; 55 end.
我要开始刷noi历年题了!我相信我一定能够坚持下来!